Question

Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.

Answer 1

Given a = 5,

Sum of n terms, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Sum of first n terms, $S_4=\frac{4}{2}[2\times 5+(4-1\left)d\right]$

$S_4=2[10+3d]=20+6d$----(1)

Sum of next 4 terms $=S_8-S_4$

So, $S_8=\frac{8}{2}[2\times 5+(8-1\left)d\right]$

$S_8=4[10+7d]=40+28d$----(2)

According to the question,

$S_4=\frac{S_8-S_4}{2}$

$2\times S_4=S_8-S_4$

$3\times S_4=S_8$

$3\times (20+6d)=40+28d$

$60+18d=40+28d$

$28d-18d=60-40$

$10d=20\Rightarrow d=2$

Common difference = 2